Examples

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

### Transcript

Question 20 (Method 1) Find the coordinates of the point where the line through the points A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane.The equation of a line passing through two points with position vectors π β & π β is π β = π β + π (π β β π β) Given the line passes through the points (π β β π β) = (5π Μ + 1π Μ + 6π Μ) β (3π Μ + 4π Μ + 1π Μ) = 2π Μ β 3π Μ + 5π Μ A (3, 4, 1) π β = 3π Μ + 4π Μ + π Μ B (5, 1, 6) π β = 5π Μ + 1π Μ + 6π Μ β΄ π β = (3π Μ + 4π Μ + 1π Μ) + π (2π Μ β 3π Μ + 5π Μ) Let the coordinates of the point where the line crosses the XY plane be (x, y, 0). So, π β = xπ Μ + yπ Μ + 0π Μ Since point crosses the plane, it will satisfy its equation Putting (2) in (1) xπ Μ + yπ Μ + 0π Μ = 3π Μ + 4π Μ + 1π Μ + 2ππ Μ β 3ππ Μ + 5ππ Μ xπ Μ + yπ Μ + 0π Μ = (3 + 2π)π Μ + (4 β 3π)π Μ + (1 + 5π)π Μ Two vectors are equal if their corresponding components are equal So, Solving 0 = 1 + 5π β΄ π = (βπ)/π So, x = 3 + π = 3 + 2 Γ (β1)/5 = 3 β 2/5 = 13/5 & y = 4 β 3π = 4 β 3 Γ (β1)/5 = 4 + 3/5 = 23/5 Therefore, the required coordinates are (ππ/π,ππ/π,π) Question 20 (Method 2) Find the coordinates of the point where the line through the points A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane.The equation of a line passing through two points A(π₯_1, π¦_1, π§_1) and B(π₯_2, π¦_2, π§_2) is (π β π_π)/(π_π β π_π ) = (π β π_π)/(π_π β π_π ) = (π β π_π)/(π_π β π_π ) Given the line passes through the points So, the equation of line is (π₯ β 3)/(5 β 3) = (π¦ β 4)/(1 β 4) = (π§ β 1)/(6 β 1) A (3, 4, 1) β΄ π₯_1= 3, π¦_1= 4, π§_1= 1 B(5, 1, 6) β΄ π₯_2 = 5, π¦_2= 1, π§_2= 6 (π₯ β 3)/2 = (π¦ β 4)/(β3) = (π§ β 1)/5 = k So, Since, the line crosses the XY plane at (x, y, 0) z = 0 5k + 1 = 0 5k = β1 β΄ k = (βπ)/π So, x = 2k + 3 = 2 Γ (β1)/5 + 3 = 3 β 2/5 = 13/5 y = β3 Γ (β1)/5 + 4 = 4 + 3/5 = 23/5 Therefore, the coordinates of the required point are (ππ/π, ππ/π, π).

Made by

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.